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给定一个由若干 0 和 1 组成的数组 A，我们最多可以将 K 个值从 0 变成 1 。返回仅包含 1 的最长（连续）子数组的长度。


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        <h1 class="title">leetcode1004.最大连续1的个数III</h1>
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            <span>二月 21, 2020</span>
            

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            <h1 id="leetcode1004-最大连续1的个数-III"><a href="#leetcode1004-最大连续1的个数-III" class="headerlink" title="leetcode1004. 最大连续1的个数 III"></a>leetcode1004. 最大连续1的个数 III</h1><hr>
<blockquote>
<p>给定一个由若干 0 和 1 组成的数组 A，我们最多可以将 K 个值从 0 变成 1 。<br>返回仅包含 1 的最长（连续）子数组的长度。</p>
</blockquote>
<blockquote>
<p>示例 1：<br>输入：A = [1,1,1,0,0,0,1,1,1,1,0], K = 2<br>输出：6<br>解释：<br>[1,1,1,0,0,<strong>1</strong>,1,1,1,1,<strong>1</strong>]<br>粗体数字从 0 翻转到 1，最长的子数组长度为 6。</p>
</blockquote>
<blockquote>
<p>示例 2：<br>输入：A = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], K = 3<br>输出：10<br>解释：<br>[0,0,1,1,<strong>1</strong>,<strong>1</strong>,1,1,1,<strong>1</strong>,1,1,0,0,0,1,1,1,1]<br>粗体数字从 0 翻转到 1，最长的子数组长度为 10。</p>
</blockquote>
<p>滑动窗口。没有用到评论区那些花里胡哨的，就是一个窗口右边进来怎么处理，左边要出去怎么处理。只不过这么写时间复杂度很高。</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">longestOnes</span><span class="params">(<span class="keyword">int</span>[] A, <span class="keyword">int</span> K)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> result=<span class="number">0</span>,i=<span class="number">0</span>,j=<span class="number">0</span>,tmp=K;</span><br><span class="line">        <span class="keyword">while</span>(j&lt;A.length) &#123;</span><br><span class="line">        	<span class="keyword">if</span>(A[j]==<span class="number">0</span>) &#123;</span><br><span class="line">        		tmp--;</span><br><span class="line">        		<span class="keyword">if</span>(tmp&lt;<span class="number">0</span>) &#123;</span><br><span class="line">        			<span class="comment">//左窗口缩小</span></span><br><span class="line">        			<span class="keyword">while</span>(A[i]!=<span class="number">0</span>) &#123;</span><br><span class="line">        				i++;</span><br><span class="line">        			&#125;</span><br><span class="line">        			i++;</span><br><span class="line">        			<span class="comment">//对于示例1来说，如果不i++，i是第一个0的位置</span></span><br><span class="line">        			<span class="comment">//但是我们想让他在第二个0的位置</span></span><br><span class="line">        			tmp++;</span><br><span class="line">        		&#125;</span><br><span class="line">        	&#125;</span><br><span class="line">        	result=Math.max(j-i+<span class="number">1</span>, result);</span><br><span class="line">        	j++;</span><br><span class="line">        &#125;</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">return</span> result;</span><br><span class="line">    &#125;</span><br></pre></td></tr></table></figure>
<p>在这之后我看了一下4ms的代码，发现差不多。我觉得唯一的差别可能在于最后的比较最大值上面。如果用max函数是9ms，if-else是8ms，而三目运算符只用了                      <strong>6</strong>ms，这个差距还是很大的。但是也不能一概而论，三目运算符不一定永远比if-else快，因为前者还可能会有类型转换的问题。<br><strong>leetcode 8/100</strong></p>

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